/*
*leetcode.com/problems/two-sum/
*Given an array of integers, return indices of the two numbers such that they add up to a specific target.
*You may assume that each input would have exactly one solution, and you may not use the same element twice.

*Example:
*Given nums = [2, 7, 11, 15], target = 9,
*Because nums[0] + nums[1] = 2 + 7 = 9,
*return [0, 1].
*/
#include <vector>
using namespace std;


class Solution {
public:
	vector<int> twoSum(vector<int>& nums, int target) {
		vector<int> res = { -1, -1 };
		int size = nums.size();
		for (int i = 0; i < size-1; ++i){
			for (int j = i + 1; j < size; ++j){
				if (nums[i] + nums[j] == target){
					res[0] = i;
					res[1] = j;
					return res;
				}
			}
		}
		return res;

		////Key is the number and value is its index in the vector.
		//unordered_map<int, int> hash;
		//vector<int> result;
		//for (int i = 0; i < numbers.size(); i++) {
		//	int numberToFind = target - numbers[i];

		//	//if numberToFind is found in map, return them
		//	if (hash.find(numberToFind) != hash.end()) {
		//		//+1 because indices are NOT zero based
		//		result.push_back(hash[numberToFind] + 1);
		//		result.push_back(i + 1);
		//		return result;
		//	}

		//	//number was not found. Put it in the map.
		//	hash[numbers[i]] = i;
		//}
		//return result;
	}
};

#include <iostream>
int main(){
	Solution sol;
	vector<int> nums = { 2, 7, 11, 15 };
	int target = 9;
	vector<int> res = sol.twoSum(nums, target);
	cout << "[" << res[0] << " " << res[1] << "]" << endl;
	int i; cin >> i;
}